Calc6.34

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Discuss the convergence or divergence of the series \sum _{{n=7}}^{{\infty }}\left(-{\frac  {1}{2}}\right)^{n}.

This is slightly different from the other series we have looked at so far, but the formulas all work even for negative values of the common ratio. So we have

\sum _{{n=7}}^{{\infty }}\left(-{\frac  {1}{2}}\right)^{n}=\sum _{{n=0}}^{{\infty }}\left(-{\frac  {1}{2}}\right)^{n}-\sum _{{n=0}}^{{6}}\left(-{\frac  {1}{2}}\right)^{n}={\frac  {1}{1+{\frac  {1}{2}}}}-{\frac  {1-\left(-{\frac  {1}{2}}\right)^{7}}{1+{\frac  {1}{2}}}}={\frac  {1}{{\frac  {3}{2}}}}-{\frac  {1+{\frac  {1}{128}}}{{\frac  {3}{2}}}}={\frac  {2}{3}}-{\frac  {2}{3}}{\frac  {129}{128}}=-{\frac  {1}{192}}\,

So this sum converges to a negative number. This might seem a bit surprising but the first term is a negative number and the terms are decreasing.


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